text:
At what point does an aircraft need a new type certificate vs a STC?
First post. Yes, I did search but this specific question has not been asked before. This answer is close, however:
How does a mechanic determine if a repair or alteration requires a Supplemental Type Certificate?
I do not own an aircraft and do not have a PPL. BUT, I love the Piaggio P180.  This question pertains to the difference between getting a new Type Certificate and getting a STC (Supplemental Type Certificate). So, which of these alterations would require a new TC of the aircraft vs a STC?

Lengthening or shortening the fuselage?
Changing the engine(s) from say piston to turbine or turbine to electric fans?
Adding extra components to the airframe, like batteries, controllers, etc. to make a hybrid aircraft?
Changing the wings to newer ones from another certified aircraft (re-winging)?
Changing the landing gear to a newer model with higher load capabilities?
Expanding the fuselage diameter by 10%?

Would implementing one of these changes require STCs or after a certain number of alterations, does the FAA require a new TC?
I'm really trying to understand when STCs are used and when the FAA requires a new Type Certificate.

Are you doing these modifications to sell kits or upgrades, or just modifying your personal aircraft? If it's personal, it will probably just require a "337", if you want to do it as a business, it needs an STC.

@Ron, changes would be for personal use.  I never heard of a "337". Will read up on it. Thank you.

Forgot to add that the aircraft would be flown under Part 23.

To start, I would suggest a read through 14 CFR Part 21.
But to summarize: a TC is necessary to manufacture a new aircraft, so TCs are the property of the manufacturer of the aircraft. The TC confers the following privileges:

§21.45   Privileges.
The holder or licensee of a type certificate for a product may—
(a) In the case of aircraft, upon compliance with §§21.173 through
  21.189, obtain airworthiness certificates;
(b) In the case of aircraft engines or propellers, obtain approval for
  installation on certificated aircraft;
(c) In the case of any product, upon compliance with subpart G of this
  part, obtain a production certificate for the type certificated
  product;
(d) Obtain approval of replacement parts for that product.
[Doc. No. 5085, 29 FR 14564, Oct. 24, 1964, as amended by Amdt. 21-92,
  74 FR 53386, Oct. 16, 2009]

OTOH, An STC can be obtained by anyone including the TC holder. The big difference is that it is a 'supplement' to a TC.  That is, it modifies the design specified by the TC.
So we can look to the following section of Part 21:

§21.113   Requirement for supplemental type certificate.
(a) If a person holds the TC for a product and alters that product by
  introducing a major change in type design that does not require an
  application for a new TC under §21.19, that person must apply to the
  FAA either for an STC, or to amend the original type certificate under
  subpart D of this part.
(b) If a person does not hold the TC for a product and alters that
  product by introducing a major change in type design that does not
  require an application for a new TC under §21.19, that person must
  apply to the FAA for an STC.
(c) The application for an STC must be made in the form and manner
  prescribed by the FAA.
[Doc. No. FAA-2006-25877, Amdt. 21-92, 74 FR 53387, Oct. 16, 2009;
  Doc. No. FAA-2018-0119, Amdt. 21-101, 83 FR 9169, Mar. 5, 2018]

As stated here, an STC applies to any major modification that doesn't require an new TC under §21.19 which states:

§21.19   Changes requiring a new type certificate.
Each person who proposes to change a product must apply for a new type
  certificate if the FAA finds that the proposed change in design,
  power, thrust, or weight is so extensive that a substantially complete
  investigation of compliance with the applicable regulations is
  required.
[Doc. No. 28903, 65 FR 36265, June 7, 2000]

So your question as to TC vs. STC really only applies to the manufacturer. And as stated here, it's at the discretion of the FAA. But as a point of reference, every Boeing 737 ever built has been built under TC Data Sheet A16WE (currently Rev 64).
Typically manufacturers will continue to modify their TC to support continuing production. They will use an STC when the changes are to modify previously manufactured aircraft.
Third parties that modify aircraft will also get STCs. A large percentage of STCs are obtained by third parties. The STCs only cover the changes from the aircraft's existing certification basis (which is the original TC plus any previously installed STCs.) An STC may be for a single aircraft, a single aircraft type, or for multiple types (an "Approved Model List STC" or AML STC). The last is common for general aviation avionics.
So back to your original list of mods. All of them are major mods and would require at least an STC, though the last one would likely require an modification of the TC if not a new one. Replacing the entire fuselage is a huge change. If Piaggio wanted to make those changes to their design going forward, they would likely modify their existing TC. If the FAA decreed the modification was extreme enough to warrant a new TC, they would have to apply for a new TC.
If someone other than Piaggio wants to make the mods, they would have to get an STC and use it to modify an aircraft that was manufactured by Piaggio. Once the STC is obtained, the STC holder has the following privileges:

§21.119   Privileges.
The holder of a supplemental type certificate may—
(a) In the case of aircraft, obtain airworthiness certificates;
(b) In the case of other products, obtain approval for installation on
  certificated aircraft; and
(c) Obtain a production certificate in accordance with the
  requirements of subpart G of this part for the change in the type
  design approved by the supplemental type certificate.
[Doc. No. 5085, 29 FR 14568, Oct. 24, 1964, as amended by Amdt. 21-92,
  74 FR 53387, Oct. 16, 2009]

One advantage to an STC, is that the owner can sell the rights to use the STC to another. So the following 'permission' typically involves the exchange of money:

§21.120   Responsibility of supplemental type certificate holders to provide written permission for alterations.
A supplemental type certificate holder who allows a person to use the
  supplemental type certificate to alter an aircraft, aircraft engine,
  or propeller must provide that person with written permission
  acceptable to the FAA.
[Doc. No. FAA-2003-14825, 71 FR 52258, Sept. 1, 2006]

I know this is an old post, but the answers here are really quite off the mark.
This is precisely the question that is addressed by 14 CFR 21.113(b), 21.19, 21.93 and FAA Advisory Circular (AC) 21.101 "Establishing the Certification Basis of Changed Aeronautical Products".

14 CFR 21.113(b): “If a person does not hold the TC for a
product and alters that product by introducing a major change in type
design that does not require an application for a new TC under §
21.19, that person must apply to the FAA for an STC.”

and

21.19: "Each person who proposes to change a product must apply for a new type certificate if the FAA finds that the proposed change in
design, power, thrust, or weight is so extensive that a substantially
complete investigation of compliance with the applicable regulations
is required."

All of them are 'major' changes per 14 CFR 21.93 and would require a TC, Amended TC,  STC, or major alteration. Any change which is 'substantial' requires a new TC. A 'significant' change, requires the modification to be certified at the latest amendment to the regulations. Appendix A provides examples as to what constitutes a 'substantial', 'significant', or 'not significant' change. Going by that Appendix, all the proposed changes, would likely be considered at least 'significant', except perhaps the landing gear change, which could be 'not significant', but is likely paired with a gross weight increase, which is likely 'significant'. The fuselage diameter change would probably be considered 'substantial'. No. 4 (re-winging) would likely be 'substantial' as well, but I know of at least one example of company (Sierra Industries) which modifies wings on Part 25 aircraft (Citation 500/550/560s) under an STC.
Finally, the FAA has also produced a Major Alteration Job Aid to help maintainers determine whether a particular change can be performed as a Major Alteration (Form 337), or requires an STC.
Of course the real question is which of these changes is viable from a commercial perspective...

FYI, the link to Major Alteration Job Aid is dead. https://www.peter2000.co.uk/aviation/sn3500/Major_Repair_Alteration_Job-Aid.pdf is a 2013 version, but is at least available.

Since this is an aircraft for personal use, is under 19,000lbs gross weight, and less than 19 seats it will fall under Part 23.
If you don't plan on selling "kits" to do these modifications, then an STC is not required. STC's are only for modifications intended for sale to aircraft for installation by an A&P mechanic. STC's are required not for the kit itself, but for each aircraft type (and sometimes variant) that it is to be installed on. This is why the STC process can be very expensive.
You would not need an entirely new type certificate either, again which would be required if you were doing these modifications as a business for sale as another aircraft type, or if the original manufacturer modified the aircraft in some significant way.
What you would need in almost all the cases you described above is an FAA Form 337, which describes "major repair & alterations to airframe or power plant". This document will require you (and your A&P) to make detailed descriptions and drawings of the modification and submit it to an FSDO. The FSDO will review the modifications and approve or disapprove of the changes. It is advisable to work with the FSDO from the beginning, not the end.

So, which of these alterations would require a new TC of the aircraft vs a STC?

Lengthening or shortening the fuselage?
Changing the engine(s) from say piston to turbine or turbine to electric fans?
Adding extra components to the airframe, like batteries, controllers, etc. to make a hybrid aircraft?
Changing the wings to newer ones from another certified aircraft (re-winging)?
Changing the landing gear to a newer model with higher load capabilities?
Expanding the fuselage diameter by 10%?

None of them require a new TC or STC, however your first question would probably be difficult at best. The second question doesn't make a lot of sense for the Piaggio since it already has turbine engines, and there are no electric fans of the size that would be required. Changing the wing or modifying the landing gear are pretty common things, usually in the form of STOL kits or FOD deflectors. Some of them would require extensive testing from the FAA before issuing the airworthiness paperwork.
The last one though will nearly be impossible. It would be a lot cheaper to just buy an aircraft that fit your needs, like the Pilatus PC-24 than try to make that modification.


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Boundary Layer Ingestion - Would it also work at supersonic speeds?
I have searched the forum and the internet but can't find the answer to my question. Would the 10% increase in fuel efficiency from BLI function at supersonic speeds?
The 10% increase is the number I read from Bauhaus Luftfahrt and their explanation of propulsive fuselage. Which is also close to the number given by NASA.  These numbers were for typical commercial subsonic aircraft and all I want to know is if BLI would also work for aircraft travelling supersonic?
Thanks, RB

Only the outer part of the boundary layer is supersonic, so the subsonic results mostly also apply at supersonic speed.

Thank you Peter.  You are a great contributor to this forum.

boundary layer ingestion was used in the air inlet duct system of the SR-71- not to reduce drag but to improve uniformity of flow across the duct cross-section. and agree with your assessment of Peter's status!

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text:
If a glider had human pedal powered propellers how much extra distance could you go?
Imagine an amateur cyclist could generate 100 watts of power (professional ones can do 200-300), would it be possible to put that to use in a glider to fly further?
For example you pedal for 15 minutes and it is stored in a battery, and then it is used for 2 minutes of propeller power.
I guess one concern is that the addition of [pedals, battery, motor] would outweigh any potential energy gains.

The possible duplicate I linked to is about a different kind of craft, but the answer's solution and reference to the MIT Daedalus project is valid here as well.

Hmm i guess i need to look at the power output of a electric glider and compare to say 100 watts to figure out how much percent it is for gliders.

Looking at ls8-e it seems like the motor is 22kw, so about 0.5% of the power, I guess it's just as impossible.

Looking at wikipedia it confirms "Sustainer engines are typically two-stroke two-cylinder air-cooled engines in the range of 18–30 hp (14–22 kW)"

From this post, it appears like:
5 lb thrust = 1 horsepower

As you didn't clearly mentioned the spec. of the glider and the pilot/passenger inside it, then will be no exact answer. But, you may take a lesson from this Albatros Gossamer. It frew 22 miles (36km) in the early morning at 12 June 1979 by just pedal power of a 55kgs its pilot, crossing British Strait. It took 2 hours 49 minutes. https://youtu.be/FfD5epGswRw

The linked answer is NOT a duplicate when it talks about autogyros.  A 40:1 glider weighing 600lb makes 15 lbs of drag, requiring roughly 3hp to sustain level flight.  If a human could produce 150 watts it works out to .2 hp, which is roughly 1lb of thrust.  This would theoretically increase the glider's L:D to about 43:1, not allowing for the drag increase from the extra weight of a drive system and propeller and extra drag of a larger cockpit pod.  So the glider could glide, at best, about 9% farther with the pilot pedaling furiously, and in reality with losses, probably closer to 4 or 5%.

Shame I shouldn't have marked it as duplicate so early. There are now two sources - Albatros flew 36km appearing to have had no assistance at the start, and also a calculation that suggests an existing glider is capable of adding 4-5% extra range.

I would reconcile them by suggesting that the albatross was probably much lighter at 32kg without pilot, versus 270kg gliders. That would make the power much more efficiently used as lift since there's less weight to keep up.

Doing a youtube search, there are several projects in the past, eg Dash PA

Details of Daedalus 88 the record holder - 100km

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text:
How are elliptical winglets better than blended winglets?
I've read in few papers that elliptical winglets give better performance than blended in terms of their L/D ratio. I am not able to find why elliptical winglets are better. the two differ by their profile followed.
Edit 1: Here  are few references
http://ijari.org/CurrentIssue/2017Volume4/IJARI-ME-17-12-148.pdf
http://www.iaeng.org/publication/WCE2015/WCE2015_pp1144-1148.pdf
Edit 2: The papers explain that elliptical winglets are better than blended just through observations from CFD study.But they do not draw conclusions on how it is better and I couldn't understand that from the contours also.

You may provide the link to the paper

Thing is its paid journal paper. I don't think I can link it. But I will try to find some other source though

@saiteja Even if you cannot link the PDF due to copyright, you can at least provide a reference to the paper (title, authors, journal, etc.).

Yes sir I have put few links

The ijari.org link is public (took me straight to the PDF). It appears to be a paper describing the characteristics of different winglets. As such, it would be a good source for an answer. What about it don't you understand?

How elliptical shows better performance...the papers just give results..it doesn't explain how and why the drag is reduced...the contours looked pointless...

The best way to understand why elliptical winglets may be better than blended winglets is to read the text of the patent obtained by the inventor of the elliptical type, Dr. Fort Felker, a former NASA wind-tunnel researcher and expert in the field of computational fluid dynamics, and co-founder of Winglet Technology LLC.
The short answer is that the curvature of the winglets as they extend outward from the intersection with the wings approximates part of an ellipse (hence their name), and this apparently ensures an elliptical distribution of lift both horizontally and vertically, resulting in lower induced drag than a blended winglet.

What do you mean by horizontal Lift distribution?...

In simple terms, the winglet is an airfoil. The vertical portion of the winglet has a lift component aligned horizontally to the wing, not vertically.


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Can time spent in consumer flight simulators increase the safety of flying light sports aircraft for licensed pilots?
I've asked this question to my flight instructor, who believes flight simulators do a very poor job of modelling flight physics for very light airplanes, which makes them useless for preventing accidents.
Considering most accidents are caused by loss of control this explanation does make sense. On the other hand, flight sims might help in dealing with engine failures.
I intend to fly in a plane derived from Zenair CH-701.

Definitely related, possible duplicate: Can Microsoft Flight Simulator help me learn to fly (or make me a better pilot)?

I agree it's related, I'm not sure whether its duplicate, as its more specific in both kind of aircraft (LSA) and aspect (safety as opposed to learning to fly in general).

I've rolled back because I am mostly interested in the safety aspect, as phrased initially.

Perhaps yes, but it's not easy to qualify objectively.
Simulating a light airplane may require a different approach in some areas, but it's not inherently difficult compared to 'heavy' ones.
However, the truth is, nobody cares that much, so there is much less investment in such simulators. A decent FTD (flight training device, something a step lower than what is legally called a 'simulator' (FFS)) will cost easily several times more than an ultralight, while the risk benefit is arguably lower (compared to real flight).
Secondly, type-rated simulators are rarity even on the 'proper' GA market; typically FTDs are 'generics' that aim to simulate a representative 'class' of aircraft. This makes fair comparison difficult, especially by real pilots who have most experience on a single/few real types.
Then, there are technical features that are even more important for light (and GA) airplanes than for big ones. One is a good visual system: in real visual flying, you obviously use it more, and having a single monitor stuck in front of you will hamper you more than you might think.
Second is control loading: on light airplanes you get airspeed and trim feedback right at your fingertips. There are no cheap solutions to this. You may argue this is not 'physics', but it actually is, this is all part of a feedback between the airplane and you being the part of control system. Lack or poor quality of these things (which is more than common on the simulators used for light planes) will substantially affect subjective and objective qualification of the flying qualities/handling of the simulator.
There are areas where even expensive FFS suffer. A typical example is ground handling. Or, speaking about 'loss of control', many regimes that result from it, notably stall and spin, are actually difficult to simulate well, and are often not required for certification.
That all said, it's an overstatement to say that simulators are 'useless for preventing accidents'. Flying is more about thinking ahead, planning and situation awareness than any specific piloting skills, and even a game simulator can teach you some of that.

This seems to me like the right answer at least intuitively. A lot of the decisions you make as a pilot don't have much to do with flight physics, but with anticipating risky situations and avoiding them in the first place. I suppose flight simulators can help in that regard.

They do a good job modeling the majority of the flight physics for airplanes with a few exceptions such as edge of envelope flight characteristics and some power plant characteristics. Eg FSX does not model stall/spin characteristics well or turboprop characteristics well or autorotation characteristics for helicopters well.  X-Plane 11 is purported to have good flight physics models in them and they can be more custom tailored to specific aircraft than FSX or Prepar3d can.  That being said, some of the aircraft handling actions are not quite the same as the real thing.
In the end desktop sims are not airplanes; they are just mathematical algorithms that closely match how an airplane flies.  It’s similar to saying that a synthesized violin sounds pretty good - but it’s never gonna be a hand made 1727 Stradivarius.  That being said those same desktop sims are good enough that the same mistakes I make in the sims, I will make in the real aircraft.
One thing anything short of a dedicated Level D full motion sim just can’t do is simulate the kinesthetic feel of an airplane.  That’s something you just don’t get sitting in an office chair with a joystick and staring at a 20” 2-D screen.  While VR sets can simulate what you see out of an airplane, it’s still not quite the same thing.

yeah, and should be added that the control feedback is different too, and nothing the sim can do to change that. Even a $1000 sim yoke can't replicate the actual control forces encountered trying to steer an actual aircraft, let alone every actual aircraft out there, let alone in every conceivable aerodynamic scenario those aircraft may encounter.

I am in solid agreement with two statements made above:

Flying is more about thinking ahead, planning and situation awareness than any specific piloting skills,
One thing anything short of a dedicated Level D full motion sim just can’t do is simulate the kinesthetic feel of an airplane.

Sims are good for practicing the patience part of flying, especially setting up for the pattern and landings, and instrument approaches and flying the procedure. Without the feedback of the plane moving with you, and just having the screen view tilt in front of you, the visual part quickly becomes unrealistic as you make maneuvers. Especially simulating the wind - the screen jumps and bounces around, but sitting in your chair it just becomes annoying.
And depending on the plane you are flying, the joystick/mouse on your computer will allow full range of unlimited movement, while depending on the phase of flight the plane may require increasing amounts of force to actually move the stick in full range, and if not properly trimmed may require a lot of pressure.  For example, I was trying out a simulator at a training school. Took off okay in a Beech Baron or something like that, but the trim was set wrong, so I had to really fight the pitch pressure while I searched for the elevator trim to keep from climbing up into a stall.

in the limit of aircraft so light that their weight is of order ~ the weight of the air they displace, the usual and customary rules of aerodynamics i.e., vehicle responses to control inputs goes out the window and the dynamical analysis gets a lot more complicated.
For example, in the case of macready's gossamer condor, deflecting the right aileron down did not roll the vehicle to the left; instead it skidded the vehicle to the right.
So it does not surprise me that consumer-grade flight simulation packages might do a poor job with predicting the response of ultra-ultralights.

You are right, but the OP's Zenair CH-701 is not quite in this category :)


text:
Companies that currently manufacture aircraft with center stick controls?
I was wondering if any general aviation manufacturers currently produce center stick control aircraft.  (Not including LSA, Sailplanes, Helicopters). I'm asking about four+ place aircraft other than the diamond da40.

I know first-hand that the Falke SF25 Motorglider -- a single-engined light aircraft with an optimal glide ratio of ~25:1 -- contains a single, centre-stick. Also, only P1 gets a "six pack" of instruments, P2 looks at the radios!

This question, by definition, is off-topic as too broad. It is asking for a list of aircraft. At the time I am writing this comment, there are two answers and both are correct. Both list at least one aircraft "still in production" which uses a center stick control. Which answer should be marked/mark-able as 'correct'? VtC as too broad, unless you can edit it to be specifically and completely answerable in a single answer.

When I edited the title I was thinking of looking at the types of planes that use center stick and why, but that is pretty close to this question.

Asking for companies that make these aircraft isn't much better, and CGCampbell's comment still applies.

Some of the lancair airplanes have it but those are kit built so Im not sure if you can count them as true production planes.

Vans aircraft
Most Variants of the Piper Cub like planes that are still made
Extra Aircraft
Pipistril Aircraft
Czech Sport (formally Piper Sport)

Its worth noting that Cessna did recently have a run trying to make the "Sky Catcher" which was quickly discontinued but center stick none the less.
A lot of light sport airplane and aerobatic planes tend to have center sticks.

Skycatcher wasnt actually a stick, more a single-bar yoke. Good list otherwise!

@egid, I did notice that but I have never flown one so I was not sure exactly how it worked although it does look like a stick ;)

Good pic of the SkyCatcher bar:
http://www.jseagle.com/wp-content/uploads/2013/09/162.jpg

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text:
Could a commercial plane like Airbus A320 glide with no working engines from 11000 meters?
I have heard that pilots learn in simulators to fly also commercial planes without working engines, so that they glide. And still I have the impression, the chances that a plane could float from 11000 meters height are nearly 0. Could someone explain, if a commercial plain CAN BE THEORETICALLY (Physics) and PRACTICALLY landed from 11km with all failed engines? How important is the altitude from which all engines fail?

It's actually better to be higher when the engines fail; I'm not sure why you think it'd be worse. With no engines, altitude=options.

All standard fixed wing aircraft are able to glide. The key measure of performance in gliding is the glide ratio which is related to the lift-to-drag ratio of the aircraft. The glide ratio of a clean A320 is 17:1 which means it can travel 17 units of distance forwards for every 1 unit of distance downward at best glide speed. Using your figure of 11,000 meters away and multiplying by 17 results in a maximum range of 187km . In practice, the aircraft would need to be higher than this (or land closer) to accomodate turns and other factors, but if the A320 was at 1000 meters AGL or higher I would expect it to have no problem landing at an airfield or airport 11km away. Note that aircraft routinely glide every time you fly even though you may be unaware of it. When a commercial flight begins its descent often power will be brought back to idle, so the aircraft is gliding at that point.
There have been several instances of large commercial aircraft landing fully unpowered. One well known example is the case of the "Gimli" glider in which a Boeing 767 ran out of fuel and glided to landing at a nearby airbase.
EDIT: It has been brought to my attention that the OP is asking about a starting altitude, not how far away from the airfield. As I imply in my answer what matters is the ratio of altitude and distance, not altitude alone.

@Tyler: Do you have a source for the 17:1 glide ratio of an A320? (Also: took the liberty to correct the answer as the question was about starting altitude of 11000m, not minimum altitude to reach 11000m distance.

Yes
On Airliners.net a participant named Longhauler posted the following in 2005

If you wish to maintain the highest altitude, you maintain your best lift drag ratio. For the A320 that's about 220 knots. You will descend about 1500 ft/minute. So from 35000 ft, it will take you roughly 25 minutes to reach the ground. You will travel roughly 90 nm.
To travel the furthest distance, you descend at M0.74/270 knots. Again, from 35000 ft, you will travel roughly 95-100 nm.
To cover a distance the quickest you descend at M0.80/330, but you be gliding like a simonized crowbar. (namely dropping like a stone). You will travel only about 65 miles, but you'll get there quicker!
As someone said above, you have to remember, pretty well all descents are done with the thrust at idle.
Another consideration, is that without engines, your cabin will depressurize. Slowly on newer aircraft, increasing with the age of the aircraft. It is a balance with finding a place to land within your gliding range, but also looking at your cabin altitude rising.

11000 meters is 5.94 nautical miles - so that's easy from cruising altitude.

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How did the 9/11 hijackers find their way to New York City?
It is quite well known that the airplane hijackers on 9/11 were not professional pilots. So, my question is how some people who didn't have any experience of navigating the airplanes could find their way, for example from Boston to New York City?
I read somewhere that some natural signs like the Hudson River (for example) helped them to find their way through New York City. But I'm still looking to find a more convincing answer if it is available.

A simple, pocket GPS is enough for that navigation, and for more complex ones too......

Comments are not for extended discussion; this conversation has been moved to chat.

The answers point out the logical flaw in the question, which is the jump from "not a professional pilot" to "people who didn't have any experience of navigating" aircraft.  In other words, "not professional" does not imply "entirely without experience."

@phoog My purpose was that at least people that we called them "professional pilots" need to have certain hours of flight in actual airplanes, and not flight simulators, to be qualified in order to navigate an airplane, which transports passengers. But at least we could say the hijackers never flew with an actual 737. For example, if they found their way from Boston to NYC so precisely to target the WTC in their first actual flying attempt, that looks remarkable from just aviation perspective (I'm not saying hijackers did a remarkable job to kill innocent people so DON'T get it wrong!).

The hardest part about flying a plane is landing. The second hardest is taking off. The hijackers avoid both parts.

Several of the hijackers, including Mohamed Atta, held at least private pilot certificates and had undergone ATP level jet training in DC9 and 737 full motion simulators in December of 2000. Atta himself held a commercial license with instrument and multi engine ratings. They were well versed in aerial navigation techniques and more than capable of navigating the aircraft in question back to New York City and Washington DC after it was hijacked and secured.
A basic scenario that any private or commercial pilot could have done:  the hijackers knew the flights they were booked on and the rough routes they would take to their destinations. A little basic planning amongst themselves would have produced the approximate position the aircraft would be at at the time it was hijacked.
Once seized and the flight crew liquidated, they could quickly determine their positions either from their headings and next waypoints or with a simple VOR fix (any competent private pilot can do this). The autopilots could quickly be disengaged and the airplanes hand flown using basic pilotage (good weather prevailed over the east coast that morning) or radio navaids to return to their targets. The navigation and flying they did that day was relatively simple.
As an update, I know these kinds of questions float around with “9/11 truthers” and other conspiracy fanatics as some sort of proof that the official explanation is incorrect. Quite often they cherry pick quotes and ignore any other evidence that won’t fit their narrative. Combine that with the public that’s largely ignorant about aviation, it allows these kinds of ideas to fester very well without challenge.

All: if anyone has any doubt about "liquidate" and its meaning please go to this question: https://english.stackexchange.com/q/467540/319476

I've left the one comment regarding the "liquidate" subject which links to english.se. Continued discussion on this point will be removed on sight.

Were the hijacked planes equipped with GPS navigation?

Maybe somebody with more knowledge of 757 or 767 cockpits could answer that question.  They were equipped with RNAV and FMS systems - that I do know - but GPS was in its infancy for civilian use at that point in time.

All of the hijacked flights were going in different directions and had to be piloted to a different destination. The hijacker pilots had different degrees of success in doing this. The flight paths they took are shown in this map published by the FBI:

Mohamed Atta, the ringleader of the effort, was a licensed commercial pilot received significant simulator training for large jets and the Boeing 737 in particular.  Marwan al-Shehhi trained with Atta and received similar simulator training for 737s. Even so, al-Shehhi apparently missed Manhatten on his first pass, although he may have just been reconnoitering his approach. The pilot of flight 77, Hani Hanjour, was also a licensed commercial pilot who had knowledge of how to operate and navigate a 737.
So, in summary, all of the hijacker pilots were trained in the basic operation of a 737 and knew how to do aeronautical navigation.

Amazing.  From this and the knowledge that the Pentagon was hit a little while after the WTC towers, the queston is raised as to why fighters weren't scrambled in time to intercept the threat coming toward DC.  Apparently according to the 9/11 commission, fighters were scrambled, but allegedly nobody briefed the pilots so they ended up flying a standard training route out over the ocean and didn't arrive in Washington until 30 minutes later (even though it'd only been roughly 13 miles away).  Sheesh.

Just to be complete, Ziad Jarrah is believed to have taken over Flight 93.  He obtained a single-engine private pilot certificate in August 2000, and had trained for multi-engine planes in a simulator as well.  Jarrah also purchased a GPS unit and some aeronautical charts shortly before the attack.

Did you draw that map yourself?  If not, it's polite to acknowledge the source (and may even be required - what's its licence for re-use?).  If possible, please link to the source as well; thanks!

I found the source - it's Public Domain, as a US Government work by FBI.

@DarrenRinger ...and so you imagine what? A bunch of F-16s converging on each hijacked plane and shooting them down?

@TylerDurden Public statements have been made that orders had been given such that had United Flight 93 not crashed, it would have been shot down by fighters prior to reaching its destination. However, you may be right in that it was unfeasable. The 9/11 Commission evidently questioned such feasibility in its report.   Tangentially related: https://www.theguardian.com/world/2010/oct/29/george-bush-thought-9-11-plane-shot-down

@DarrenRinger -- two F-16s were in fact on course to intercept Flight 93.  They had been given orders to shoot down the airliner but no ammunition with which to do so.  The two pilots, Lt. Heather “Lucky” Penney and Col. Marc Sasseville, discussed it en route and agreed to try to ram the much larger aircraft, if it came to that. Penney's father was a pilot for United; she knew it could very well have been his flight she was bringing down.

The 9/11 Commission Report goes into some detail on the hijackers' planning and preparation, including a (not entirely successful) attempt to obtain aviation GPS units:

Moussaoui  also  purchased  two  knives  and inquired  of  two  manufacturers  of  GPS  equipment  whether  their  products could be
  converted for aeronautical use
[...]
On August 22, moreover, Jarrah attempted to purchase four GPS units
  from a pilot shop in Miami. He was able to buy only one unit, which he
  picked up a few days later when he also purchased three aeronautical
  charts (page 247-249)

The report doesn't detail whether there's any indication that unit was actually used. However, during their flight training, two of the hijackers also took a number of practice flights that would have familiarized themselves with the areas around New York and DC, which could have helped them with visual landmarks:

Jarrah and Hanjour also received additional training and practice
  flights in the early summer.A few days before departing on his
  cross-country test flight, Jarrah flew from Fort Lauderdale to
  Philadelphia, where he trained at Hortman Aviation and asked to fly
  the Hudson Corridor, a low-altitude “hallway” along the Hudson River
  that passes New York landmarks like the World Trade Center. Heavy
  traffic in the area can make the corridor a dangerous route for an
  inexperienced pilot. Because Hortman deemed Jarrah unfit to fly solo,
  he could fly this route only with an instructor.
Hanjour, too, requested to fly the Hudson Corridor about this same
  time, at Air Fleet Training Systems in Teterboro, New Jersey, where he
  started receiving ground instruction soon after settling in the area
  with Hazmi. Hanjour flew the Hudson Corridor, but his instructor
  declined a second request because of what he considered Hanjour’s poor
  piloting skills. Shortly thereafter, Hanjour switched to Caldwell
  Flight Academy in Fairfield, New Jersey, where he rented small
  aircraft on several occasions during June and July. In one such
  instance on  July  20, Hanjour—likely  accompanied  by  Hazmi—rented
  a  plane  from Caldwell and took a practice flight from Fairfield to
  Gaithersburg, Maryland, a route that would have allowed them to fly
  near Washington, D.C. (page 242)

The report also says that several of the hijackers also had access to flight simulator software and/or simulator time at flight schools, which would have given them further opportunities to practice navigation.

I like this answer cause it tries to show the evidences based on a reference.

The Hudson River will take you straight to Manhattan. On a clear day, such as September 11 was, you could see the WTC for 100 miles or more (I had seen it from 160 miles away at Montauk Point on clear days, but you had to know where to look).
AA77 would have had a problem navigating eastbound with such an amateur as Hani (who most likely never had the certificates reported). Since there aren't any prominent landmarks going east from Falmouth VOR.


text:
How can flying high minimize drag and increase fuel efficiency?
As a plane goes higher, the engines have to work harder to compensate for the air density, therefore it will require more fuel in order to provide the same power at lower altitudes.
But I always hear that flying high means less fuel burnt.
How can that be the case?

Less air molecules means less drag...

On what basis do the engines have to "work harder"?

Flying at high altitude produces the same amount of drag, because you must fly faster in order to get enough lift power. BUT, it allows you to cover more ground miles with the same power setting in given amount of time, compared to what would you cover flying at low levels. Why? Because the air density is lower at altitude. There's also a significant temperature benefit for the engines. They run more efficient in low temperatures.

Related, maybe even a dupe?

Greg, I flew multi engine airplane and I've noticed that for every thousand feet my RPM indicators slightly decrease. I have to add more power to maintain the same airspeed. some times it get to the point where I'm at full power and I'm barely getting the normal cruise speed.

I thought (internal combustion) engines were most efficient at WOT, and higher altitudes allows that

This question doesn't make sense unless you specify what kind of engine you are talking about -- reciprocating, turbofan or turboprop.

@Xen2050 thermodynamically most efficient, possibly. You may see a thrust hook showing that TSFC ('efficiency' of a sort) not perfectly decrease with throttle setting (since lower TSFC is better 'efficiency').

Flying higher means less drag because the air is thinner; therefore, you can fly faster at altitude and hence travel farther on less fuel.
However, flying higher also means less oxygen available to burn your fuel, so available horsepower decreases with altitude.
There is a special altitude at which these two effects (drag reduction and available power reduction) achieve a crossover point. This is called the critical altitude and it is there where you achieve optimum cruise and best economy.
If you want to fly higher than critical altitude, you will fly slower because the power loss is greater than the drag reduction and you will spend more time in the air.

You don't necessarily fly slower above that altitude, you just fly more inefficiently. The issue is that the stall speed increases at the same point you start approaching critical mach, so the "window" between these two speeds gets smaller, which is known as the coffin corner.

Got it, thanks. Is it true that above critical altitude, to maintain altitude constant requires progressively larger amounts of pitch-up trim? if so, is the inefficiency issue related to increased induced drag?

In fact the aircraft is designed to fly at that sweet spot and ideally it would only fly at that altitude and speed. The design problem is to get the aircraft to fly for takeoff and landing speeds. Hence flaps.

"so available horsepower decreases with altitude"  I would stress this part - because if you want  and can obtain that up to  ... then you will save fuel all the way up to that height with limited tradeoffs to power.  Above that - you start trading power for drag.

If you read other answers, it appears that the thrust reduction is because of the reduction of available mass rather than o2

There is not 'a special altitude' - there are various combinations of altitude and throttle depending on what you want to optimise for - distance, payload, loiter time for observation etc, and what configuration the jet is in - inclusion of external tanks, pods, missiles etc etc etc.  I'd expect even civilian jets vary depending on minimum fuel or minimum journey time and different weights.

sorry, I assumed a normally aspirated piston engine driving a propeller. in that case, the critical altitude lies between 7500 and 10,500 feet, depending on stuff.

Adding to the answer by Niels Nielsen, the special altitude changes (increases) during the flight as the power required reduces as the weight of the aircraft decreases due to fuel consumption. But these effects also depend on the type of propulsion (jet vs propeller). Basically, for propeller aircraft the plot should be power required and for jet aircraft it should be drag. Now, you draw a line tangent to the curve going through the origin. The point where the two lines meet is the optimal velocity for maximum range. Power required will shift down and to the left for decreasing weight and up and to the right for increasing altitude. The drag curve will approximately translate to the right for increasing altitude.

The approximate analytical equation for range is given by the Breguet equation:
$R_{prop}=\frac{\eta_j}{c_P}\frac{C_L}{C_D}ln\frac{W_1}{W_2}$
$R_{jet}=\frac{V}{c_T}\frac{C_L}{C_D}ln\frac{W_1}{W_2}$
$\eta_j$ is the propeller efficiency and $c_P$ and $c_T$ are the specific fuel consumptions, which remain approximately constant regardless of airspeed and altitude changes (relative to cruise altitude). But it does decrease with altitude. Propellers are also less effective at high altitude and also have a sweet spot similar to jet engines. Moreover, the required lift should be equal to the weight for a given time instant, which is given by:
$L = W = C_L\frac{1}{2}\rho V^2 S$
where $\rho$ is the air density and S is the wing surface area.
Now to maintain the condition for maximum range and fly at a constant $\frac{C_L}{C_D}$ either the air density must decrease or the velocity must decrease. Decreasing the velocity is not preferred since the aircraft will fly slower and slower meaning longer flights. This flight profile is often referred to as cruise-climb flight.
A great book that treats a lot of these types of aircraft performance considerations is:
Elements of airplane performance by Ger J.J. Ruijgrok

I see you cannot comment but if you improve your answer to show how maximize efficience for each type of engine I'm sure you wiull get enough upvotes to allow you to comment in the future

Adding on Ralph's comment: answers here should be self-contained. Adding and explaining one of these power required/available plots to your answer could achieve this.

This is a good question but first we must clarify something.
There is a major difference between propeller driven aircraft and pass through turbine propelled aircraft. Without getting to technical in this explanation, let’s leave propeller driven aircraft alone for the moment and keep our answer to the confines of high bypass ratio turbine driven engines.
The long definitive answer involves a great deal of physics, another subject we will not get to rhetorical about at this time. We will however consider Bernoulli’s principles concerning fluids and the airfoil or wing. This brings us to our first thought; consider the atmosphere, or air, around us as a fluid. Now that was simple enough.
When an airplane takes flight it must defeat its greatest foe, Newton’s contribution to this question, gravity. In its most simple terms, the ability to defeat gravity requires an opposing force that is greater than the weight of the mass being lifted. In aviation this comes from the lift created, of course, by the wings.
To help make this possible on airplanes the wings are shaped with a leading edge, that is the part that first hits the air, then a highly rounded upper surface, known as camber, and much flatter lower surface which come together at the trailing edge. (Just a moment more, I’m getting to your answer.) Now we apply Bernoulli’s principal.
Looking at air as a fluid, in principle Bernoulli’s theory is this. The air traveling over the top of the wing is moving at a faster speed than the air traveling under the wing. In theory, this would cause greater pressure under the wing (lift) and less pressure over the wing as the air is rushing from the leading edge to the trailing edge. The reasoning here is that when the air molecules are separated by the leading edge of the wing, the molecules passing over the top, which has a longer surface due to the added camber, must move faster in order to reach the trailing edge at the same moment as the air passing below the wing. Now we get to the part concerning thicker (heavier) air mass and thinner (lighter) air mass and the effects on “lift”.
In order to for the wings to produce the lift needed to defeat gravity you must have thrust and lots of it! Now we enter into the power (engines) production that is required to create flight. As we all know, an airplane must be moving into the air mass in order to obtain the effects required to produce flight also known as the lighter than air effect. So in principle, the shortened version, the air passing over and under the wings must produce a lift equivalency greater than the weight of the aircraft. There is a formula but we’ll skip that here and just state this, the airmass flowing around the wings must be equal to or greater than the mass of the airplane. There, we have flight... well sort of.
You see the weight of air at sea level is much greater than the air at say 30,000 feet. Here again, in principle, the amount of air flowing across the wing on takeoff, by mass, is more at mean sea level than it would be at 5,000 feet. Since this is a principle of mass verses mass (in this case in order to defeat the effects of gravity), the fluid principles of Bernoulli’s would apply well here. So what are the effects on engine performance?
Modern power plants today are engineered to direct less airflow through the engine or turbine section of the power plant. The ratios have changed dramatically in the last decade with most engines diverting 80% or more of the air flow around the core and exiting as additional thrust at the tail. These engines, as one could imagine, consume extreme amounts of air as it flows through and around the core, or power unit, and exits the tail as super heated energy and thrust. To maintain this thrust the amount of air mass must remain fairly constant. So, how is this accomplished at 38,000 feet? And what about the thinner air passing over the airfoils?
Well, here is where physics and the dynamics of air mass come into play but in a much simpler way than you might expect. Remember the role that the wing plays in all of this? Air pressures above and below the wings can still be easily maintained at higher altitudes. Why? Because of the much greater amount of air passing over the wing surfaces at speeds often over 500 mph. You see, it’s the air molecules that hold the secret to air mass and density. At speeds like this the air density is super compressed to allow far more lift than needed to maintain flight. Then there is the engine question.
Since air is composed of gasses (mainly nitrogen with about 20% oxygen and other gasses mixed in) at the molecular level, they too can be “squeezed” under tremendous pressure and therefore compressed to create an even mixture through a jet engine. This ability to compress huge amounts of air while flying at high altitudes again makes up for loss of mass and density. Since the engine is moving between 500 and 600 mph, the intake or ingesting of plenty of air through the core and surround is no problem.
The added bonus is this; thin air creates less friction on the aircraft and much cooler air temperatures (normally around -60 degrees) as the air flows through the engine. There is also a slight reduction in fuel cost flying at these higher altitudes as we normally reduce our throttles 10 to 20 percent while at cruising altitude. This of course is dependent on the aircraft type and any head or tail wind!

the molecules passing over the top, which has a longer surface due to the added camber, must move faster in order to reach the trailing edge at the same moment as the air passing below the wing.  This is the classic "equal transit" fallacy. Please see NASA website for further explanation.

Furthermore, "at speeds like this the air density is super compressed" - not true. In fact, in most flight regimes compressibility is almost negligible. Overall, most of these explanations (including about the engine, throttles, etc.) are not only wrong, but irrelevant. The actual answer is simpler: at altitude, to have the same required lift (exactly, and not "greater", as the weight) you must fly faster due to lower air density, while having about the same drag and thus thrust (depending on wing optimisation). Finally, you don't need engines to "create flight": just look at gliders!

First, “equal transit” and Bernoulli’s principle. Really? Let’s just change all the dynamics of fluid flow. Second, I just fly them. Big ones. I wouldn’t understand anything about flight yet my airline puts me at the controls of something I don’t understand. There is a factor of air compression while at high altitude flight. Without it we would starve for oxygen. Our throttles, though automated while on AT, are adjusted based on head and tail wind. Drag, weight, lift, thrust, cord, deflection, angle of attack, lift coefficient, etc. Are we in flight school? What gets the glider up there?

Joe, please don't get offended. Pilots don't need to understand (and rarely do) all the details of aerodynamics. But "equal transit" is such a wrong explanation of Bernoulli principle that it's beyond the lie to children stage. The length of the top and bottom has nothing to do with lift. How would a flat plane create lift? As for compression, engines compress air for themselves (hence the need for turbine) and "we" get it from them. But aerodynamic compression is not a significant factor until about M>0.6, and depends only on M, not altitude.

Zeus, I started my career at Pan Am as a flight engineer on a B-727. That was over thirty years ago. We were required to have a very good understanding of flight. Not so today. Since that time NASA has “re-engineered” the theory of powered flight. “A”, “B” and “C” no longer apply. The four constants still remain; drag, weight, thrust and lift. As I extend my leading edges and flaps on the 777, I add 30% more surface to my wings to compensate for a greater attack angle at lower speeds. At cruising, the wings are engineered to reduce drag but maintain lift. Aerodynamic center is relatively new.

Zeus, my point is that equal transit, molecules separated at the leading edge will travel at different velocities and end up next to each other at the trailing edge, has been disproved. However, the effect, as theorized by Bernoulli, that the pressure above the wing is less than that below the wing is correct. So with my flaps fully extended, some of his theory is at work. But as I rotate my nose, I change my angle of attack while at the same time increasing pressure under the wings. Aerodynamic center doesn’t change. With Newton’s second law applied here we create powered flight. Thanks, Joe


text:
How do aircraft land on snow?
It is very common today for planes to land on snow (for example, in Antarctica). How are planes kept in control with so little friction against the ground? What is the procedure to perform this kind of landing? How much extra distance, on average, does an aircraft cover by landing on snow, compared to landing on dry pavement? (Video)

4 US Air Force Hercules aircraft at Willies field, Antarctica Source

It is very common for planes to land on snow (i.e. packed snow) or even ice runways.

How are planes kept in control with so less friction on ground?

Snow/ice friction is a function of the outside air temperature, so it depends. Another factor is whether the runway was packed, fresh snow - and the temperature the previous day. The colder it is, the more friction the snow will have (i.e. adequate braking). In warmer temperatures the snow will be slick (scarily slick at times - as in nil braking).
While flying a multi-engine airplane, one can use differential power. If you're flying a single-engine airplane - one is better off with small taps on the brake during slick conditions. If the OAT is relatively cold (well below freezing), braking can be pretty good.

What is the procedure to perform such kind of landing?

The procedure is the same as a standard landing. However, it is awfully nice to have good contrast on the snow to see the runway. Often runways that are pure snow/ice will be more visible than the surrounding terrain. At the very least - the runway will be outlined with some sort of runway markers.

And, on an average how much extra distance does an aircraft
  covers by landing on snow, as compared to landing on ground?

As I said before, it is really dependent upon the outside air temperature and the condition of the runway (i.e. packed snow or fresh snow).
For example, suppose there were fresh snow on a relatively warm day (30 degrees farenheight) and the snow is compacted (i.e. by humans) that day. Say, night-time temperatues drop to 0F, and the next day you land with an OAT of 30 to 35+F... that runway is going to be SLICK. However, if the temperatures were to stay at 0F - you would actually have decent braking. Even further, if the OAT dropped to -40 below...you will have excellent braking.
TLDR:
Snow conditions are always changing and braking/control of the airplane are dependent on those conditions. Snow-conditions are highly dependent on the outside air temperature and to some degree human-maintenance (for better or worse).
*source: Personal experience from flying in the Arctic.

"Often runways that are pure snow/ice will be more visisble than the surrounding terrain."—runways are usually pure snow/ice because the terrain is completely snow/ice covered, in which case everything is white and you need good markings.

+1 for mentioning visibility problems. If all land below is white, it is hard to estimate altitude.

I'd quote from this source

Landing on snow itself is not a problem. The rudder is used for steering upon landing so control is no different than landing on a dry runway.
Getting stopped is where it can get alittle tricky. Careful use of reverse thrust and braking along with not making any turnoffs from the runway until sufficiently slowed are the name of the game.
A wise Capt once told me not to make any turns offs from a snowy/slippery runway until your slow enough that you have to actually add power to get off. Wise advice indeed.

and

Landing on wet or contaminated runways requires more
close control over approach speeds.
With a B747 you should accomplish touchdown on the
runway at 1.000 ft from the approach end of the runway. The airplane should be flown firmly onto the
runway at the aiming point. It's very important not to land long! The speedbrakes should be used immediately
after the main gear contacts the runway, because they
destroy lift, increase drag and, most important, increase
main gear loading. It's necessary to use thrust reversers at high power as soon as possible because
they are most effective at high speed.

Hope this answers your questions. TY

I saw that link before asking this question, and was not convinced with the answers. Just see  how complete and thorough is @fbynite's answer.And, the third post there is specifically for the B747-200 only.

I only fly single engine aircraft.  I have landed a number of time on snow and on ice.  The two things I do differently is to 1) feather the touch-down with a slight power increase just before hand, and 2) "fly" the aircraft all the way to a near stop.  That is to say that instead of steering the plane once on the ground, you treat it more like you are still in the air.


text:
Which company manufactures the air data system on the Boeing 787-9 'Dreamliner'
I'm asking specifically about the manufacturer and model of the Pitot probes/tubes that provide the pilots with current air speed readings.
Pretty sure it's either UTC or Honeywell but can't find anything definitive online.
P.S - Is there any evidence of this technology being improved? Especially since Air France flight 447, and more recently Lion Air 610, where in both cases the pilots lost control of the aircraft due to inaccurate airspeed measurements.

There is zero evidence at this time that the Lion Air flight had inaccurate airspeed measurements. Only irresponsible people speculating.

Oh is /r/aviation spilling over here?

For the purposes of this discussion it’s safe for us to entertain the likelihood that this accident was in part caused by inaccurate airspeed readings.

"in both cases the pilots lost control of the aircraft due to inaccurate airspeed measurements". For AF447 this is a simplification. Most of the time during the event, speed indication was reliable. Transient unreliable airspeed is something the crews are trained to identify and deal with. Lack of proper training and inefficient cockpit management are more the causes of improper manual mode reactions which led to the tragedy, as pointed out by BEA in their report.

@mins I completely agree.

Boeing procured the avionics for the B787 in "packages". Air data was part of the "Navigation Package" which included in addition to air data, IRUs, nav radios, GPS sensors, and the FMS software.
Honeywell was selected as the supplier for the Nav package. So they provided the air data system.  I don't know what specific system they supplied.


text:
Can quadcopters still fly with propellers at an angle?
Can quadcopters still fly with propellers at an angle? By this I mean like the whole motor mounts will be situated along a certain slope (so as to be neither horizontal nor vertical).
UPDATE
I meant something like this (I know it's a crude diagram, but I hope it gets my meaning across better):
I don't think that's how most drones fly, if the surfaces the motors are on are folded in a way that resembles a flying motion of birds or butterflies.

Isn't that how they go forward normally?

There wouldn't be a reason they wouldn't "Fly" still. If you had them all orientated on a slope forward, depending on the angle, it would fly with a always forward motion. You wouldn't be able to hover as the thrust is always pushing the aircraft in one direction or the other.

@quiet flyer Ah, let me edit the question a little bit...

So instead of the rotor shafts all being parallel to one another, with the blades all spinning in the same flat plane, you'd them leaning out in pairs, or all leaning out from a central imaginary point below the craft, something like that?  I can't tell if your picture is intended to be a V, so the blade moves air down and out, or more like the roof of a house, the air moves down and in.

Why would you want to angle the propellers in such a manner? Without a really good (and sound) reason, such a configuration simply wastes energy because the lateral components cancel out, while usable lift is reduced.

@CrossRoads more like a V

I didn't see any on Amazon that were anything besides in a flat plane.  I would think it probably makes self stabilizing control harder.

Yes, you can. But it very depend with the angle (as in your drawing). If it is "folded) to make V-shape (that will make divergent thrust), then will reduce the lift compared to if the angle is opposite direction (inverted V-shape), which will make the thrust become convergent, it will produce more thrust. Both will work, but one is better than another. See this Akagi Helicopter. But the angle should not be high enough.
There is research that researching eagle and bird shape. There are two kind of bird or eagle's shapes. The 1st is the wing feathers trailing edge is divergent, which is make it look like V seen from the head. The 2nd is, the wing feathers trailing edge is convergent (closing to the tail, so it look like V seen from the tail, not from the head). From the research, the 2nd is better than the 1st in maneuvering and speed. But in slow move (like surveillance), the 1st is better as the flaps are produce less speed and power for the same amount of flapping (power). Another thing, the power of the flapping of the 1st is divergent, leaving the bird/eagle body, while the power of the flapping of the 2nd eagle is concentrated beneath the body (convergent) that make amount of lifting power.  Unfortunately I can not find the video in Youtube. Maybe later.

Hello, thank you for all those details. Just to clarify, in your 2nd paragraph, the "1st" is te V shape and the "2nd" is Ʌ shape — is that right?

Also, could you provide any sources you referred to in writing that? I'd like to do some more research on my own. Thank you :)

Hi, here is the video. Seems I have mesh-up my explanation. But clearly you can watch the complete explanation in this video. Feather Flight - How do birds fly? https://www.youtube.com/watch?v=1kztP_XTEPk

So, have the propellers mounted at an angle to each other? Sure, it'll still fly.
Normally, the propellers of a quadcopter all produce thrust in the same direction, and all of the thrust contributes to holding the quadcopter up (and to rolling and pitching it, depending on how fast the propellers are going relative to each other).
If you angle the propellers into a V or something, then the propellers will produce some thrust in concert with each other and some thrust in opposition to each other. So, part of the thrust will contribute to holding the quadcopter up, and part of the thrust will serve absolutely no purpose at all.
The result: the quadcopter will still fly, but it will waste some energy as it does. I'm not aware of any advantages to having the propellers angled like that.

There is also the cost aspect to consider.
It is far less expensive to alter the speed of the propellors than to tilt them, which involves heavier, swiveling mounts, plus extra motors or servos to swivel the propellers and motors, and more power needed to lift the extra weight and counter the loss of lift.
Note that on aircraft,  propellers are always perpendicular to the direction of travel, to maximize the thrust they produce. Even true of tilt rotors, which typically operate with the rotors either vertical or horizontal, except when in transition from hover to forward flight and back.


text:
Is it possible to fly to any destination on a rhumb line track?
If we follow a constant rhumb line track, it would eventually lead to the pole (depending on your hemisphere) , so then it may not be possible to fly between any two points on a rhumb line track?

Looking at the picture it gives the impression that I would reach the north pole before I would be able to reach my destination

why not? 2. this is more of a geometry question, than aviation.

I do not think a Rhumb line that follows a Latitude will ever lead to a Pole.   The most basic example is circling the equator by going Due East/West.

Sure you could.  You could even fly a Fibonacci spiral to your destination if you like, but why?

To all asking, please look at the picture to know why I have this confusion

@SachinChaudhary I'm not sure how that picture helps. It doesn't show the destination you want to fly to! The fact that most rhumb lines eventually lead to a pole seems to have nothing to do with your actual question.

This question is effectively asking that if you keep moving towards the edge of a table, will you eventually fall off? Well, yes, because the path you describe by definition keeps moving towards a pole at a constant rate.

What the picture demonstrates is that if you choose a direction to fly without regard to where you want to go, you will likely not reach your desired destination.

If you flew the rhumb line you depicted, you would fly through each and every point along the track.  So why do you say it isn't possible to fly through two points along it?  And if you are talking about two different points not on this particularly rhumb line, then simply draw a new rhumb line that connects them.  You haven't done a very good job at explaining the basis of your confusion, and I don't know why this question has 6 upvotes...

A rhumb line is a line which crosses all meridians at the same angle. Such a route is of constant true heading. This property is the reason rhumb line navigation was invented, and Mercator maps created.
On a Mercator map you just draw the straight line between the two points. This means you can always draw a route between two arbitrary places. E.g. let's connect:

Rio de Janeiro
Nome, a small city in Alaska
An abitrary point at coordinates lat = 85°, lon = 0°.
Hawaii
Stonehenge

Dotted lines show how the route completes pole-to-pole (full rhumb lines)

On most projections, rhumb lines appear as curves spiraling between poles, e.g. on this orthographic projection:

Why rhumb line navigation in the first place?
16th century navigators knew perfectly the shortest route was a great circle. But following a great circle requires constant and precise heading adjustments.
To simplify navigation, they split a great circle route into segments, and navigated each segment as a rhumb line, that is a route within the same rhumb, a rhumb being 1/32th of compass rose (a 32-point rose was often part of the map). They adjusted the heading only at the beginning of the next segment, possibly after they confirmed their position using a sextant.
They used maps just invented by Gerardus Mercator for the purpose of facilitating drawing rhumb lines and determining headings.
Aircraft with computerized navigation systems and autopilots are able to adjust the heading as required. They restored the interest in pure great circle navigation.

Please look at the photo in my edit. Looking at the picture it gives the impression that I would reach the north pole before I would be able to reach my destination.
Or is the case that the rhumb line atleast first would lead me to my destination, and after that if I continue on that track I would go to north pole?

@SachinChaudhary The latter. If you start close to the equator and head nearly due east (say, 89°), you will circle the Earth multiple times before getting close to the pole. (Drawn on a map in the Mercator projection, your route will comprise multiple parallel lines stretching all the way across the map).

As long as your points of departure and destination are on the same hemisphere, you can connect any such points with a rhumb line. It's just a matter of selecting a suitable angle with regard to the longitudes.

@Jpe61 -- all the posted answers seem to show that the departure and destination do not need to be in the same hemisphere

@quietflyer: Actually any rhumb line goes from one pole to the other, and connect any pair of intermediate points regardless of their latitudes.

@mins Except for rhumb lines going straight east ($\beta=90^\circ$) or west ($\beta=-90^\circ$). They'll never reach a pole since they maintain latitude. The term loxodrome explicitly excludes these special cases.

@Bianfable: Yes, "any" except these cases.

@SachinChaudhary: “Looking at the picture it gives the impression that I would reach the north pole before I would be able to reach my destination”. Why do you say so? Which is your destination? The curve in your image touches a lot of points (infinitely many!). If your destination isn't among them, you have chosen a wrong angle beta; just chose the correct one, according to what this answer explains.

Holy crap you are right @mins, of course. Now that I actually think about it, it's obvious. Not always wise to go by memory...

Yes, it's possible to fly from any point on the Earth from any other point on the Earth via a Rhumb line. The destination pole depends on the course you select, not the hemisphere you happen to start in. Just because a Rhumb Line leads to a pole doesn't mean it leads to the nearest pole. A track of, say, 135 is always going to terminate at the South pole, even if you start in the Northern hemisphere.

If you fly an orthodromic course, the rhumb changes continuously during the flight unless the points of departure and arrival are in the equator or in any great circle that passes through the poles...
It is, however, always possible to fly a constant-rhumb course –a loxodrome– between any two points on the planet.


text:
How are degree of freedom and gimbals related in a gyro?
Why does it mean to say a gyro has an even degree of freedom?
Why does this degree correspond to the number of gimbals ?

A gyroscope can move freely about its spin axis, this equates to a single plane of movement but is not defined as a degree of freedom.  A gimbal allows movement around another plane in addition to the spin axis, this is called one degree of freedom.  The number of degrees of freedom equals the number of gimbals.

Thank you so much.


text:
Is QDM displayed wrong for the VOR in the aircraft?
Clearly, the VOR signals follow a great circle track. Since, they follow great circle tracks, the track to the station, measured at the aircraft (either, true or magnetic), would be different due to convergency. But the QDM display doesn't take account the earth convergency, it is simply reciprocal of the QDR. So isn't the information displayed slightly incorrect?
Or, is QDM supposed to be a rhumb line track to the vor? in that case we need to apply conversion angle .
SO QDM = QDR +/- 180 would turn out to be wrong right?

Does this answer your question? Is QDR = QDM ± 180 always true in the case of an NDB?

no, that is in a different context

How so?  (besides NDB vs VOR, the answer is the same, no?)

@MichaelHall closing questions based on the answer being the same is a bit problematic. They are similar questions though, I agree.

@Jamiec, Beyond a vague similarity, the fundamental concept behind the two questions appears identical.  If the OP was responsive and could clarify their confusion over possible perceived differences I would be understanding.  Otherwise, should we expect another question about the effects of earth curvature on TACAN bearings a year from now?

@MichaelHall I take your point entirely. I'll reserve VtC but if someone else does i'll not object at all.

Maybe we should edit the first question to be more broad and cover any ground based navaid?

While you're technically right, that VOR radials are great circles, the effective range of a VOR makes the difference from magnetic whilst tracking a radial irrelevant.
So, for all practical purposes, QDM does indeed equal QDR +/- 180 degrees.

Thank you. Also, Is QDM "supposed" to be the magnetic "rhumb" line track to the station,
or is it supposed to be the magnetic "great circle" track to the station?

@SachinChaudhary Great circle


text:
Could a quadcopter fly if its propellers were encased in cylindrical tubes?
Picture a standard quadcopter, with regular old propeller guards, which are generally just circles around the propellers. Now, take the propeller guards and stretch them so that they're cylinders surrounding the immediate cross-section of propeller blades, as well as some distance above/beneath them.
Now I am fairly certain that it would be able to go up/down. But would it be able to turn? I know it is a relatively vague question, but after thinking about it for some time, I really have no idea. I might just modify the propeller guards on my cheap little drone to see if this works.

Place to start reading: https://en.wikipedia.org/wiki/Ducted_fan

Oh nice, I didn't know that was what they were called, thanks! I'm trying to make it such that the cylinders are longer than what a standard ducted fan looks like (it seems like they only protrude past the propeller a bit) but this is a good starting point. Thanks!

Yes. It would be able to turn. Of your 4 propellers, 2 spin clockwise and 2 spin counterclockwise.  The torques cancel out.  To make it spin, you run the 2 clockwise propellers a little faster and the other 2 a little slower.  Now the torques do not cancel out and the quadcopter spins.  This does not change if you have a ducted fan versus a propeller

Yeah that's kind of what I was thinking, is that it is virtually the same system so it should work the same. The thing is my design requires "ducts" that are approximately 6 inches long so I'm not sure how efficiently it would be able to achieve static flight. I know ducted fans are less efficient when the vehicle is not in motion... Thanks though!

Yes, such a craft would be controllable in all 6° of freedom. What you have essentially designed there is a quad copter powered by ducted fans as opposed to traditional propellers. The concept still works, regardless.
Quadcopters control yaw by means of reducing motor power to one set of transverse rotors and increasing power to the other set up. The upshot of which is the aircraft has the same amount of lift but an imbalance in net torque about the vertical axis, causing the craft to yaw.  The craft you suggest could be control about its vertical axis using a similar system, or you could do so using an auxiliary ducted fan to create the imbalance in torque about the vertical axis.

Yes, it could fly.  It would yaw and ascend and descend as usual, but the extra mass of the ducts would slow down how those maneuvers start and stop.  (It would feel sluggish.)  Ducts help when diameter cannot be large and efflux speed matters more than raw thrust.  But thrust is what's needed for hovering by any kind of copter, quad or not, and thrust depends strongly on diameter.
On a quadcopter, ducts might help in a straight-line high speed drag race, but not for more general missions that need endurance or agility or payload.

Yes, they'd essentially be thrust nozzles.

Can you give a bit more explanation about why that would be the case? On Stackexchange it's encouraged to explain your answer instead of one-liners :)


text:
Why $\frac{\Delta p}{\overline q}$ is depending by weight, flap deflection and other parameters? Which are other parameters?
As you can see in this link (Roskam, Airplane Aerodynamics and Performance, Chapter 2.5.3 Airspeed corrections) ...
... in the first and second line of page 30 Roskam tells that, in hypothesis of incompressible flow $$ \frac{\Delta p}{\overline q}=\frac {\frac{1}{2}\rho_0V_c^2-\frac{1}{2}\rho_0V_I^2}{\frac{1}{2}\rho_0V_c^2} $$ " depends on factor such as weight, flap deflection and other airplane parameters ". Why does it depend on weight, flap deflection?
Which "other airplane parameters" do affect $\frac{\Delta p}{\overline q}$ ?
Note: for completeness of definitions, above in the same link, page 27  Roskam is defining:
"[...]  $V_I$ (IAS) is the indicated airspeed corrected for instrument error, $\Delta V_i$, only:
$V_I=V_i+\Delta V_i$      ________________$(2.54)$
$V_c$ (CAS) is the calibrated airspeed. It is equal to the indicator reading corrected for position error due to incorrect static port location $(\Delta V_p)$ and also corrected for instrument error:
$V_C=V_I+\Delta V_p$      ________________$(2.55)$ [...]"

Sorry, but the linked page will only show by chance - in my case, "La pagina non è disponibile". Without that context I would need to guess what the question is about.

The formulas deal with the relationshop between calibrated airspeed and indicated airspeed. The difference is due to two factors:

instrument errors: these errors occur within the airspeed indicator instrument when transforming the pitot pressure to a speed indication.
position errors: these errors are caused by the position of the pitot tube on the airframe causing the pitot pressure from being different than the freestream pitot pressure.

Since the pitot tube is usually close to the fuselage of the aircraft, the local airflow around the pitot tube is influenced by the aerodynamic properties of the aircraft itselfs. The pitot tube's location is chosen such the the local flow characteristics deviate as little as possible from the freestream characteristics but there will be a measurable difference in pressure between the freestream total pressure and the pitot pressure. The pressure differences depends on airflow around the aircraft which is infuenced by the angle of attack (and thus weight) flaps, slats, gear, turn rate, side slip angle etc.
In order to analyse the position errors during a flight test programme, test aircraft are often fitted with additional pitot probes protruding ahead the airframe and/or trailing cones (for static pressure measurement) far behind the aircraft.

Thank you. Does someone know how is it affected by weight, too?

The weight affects the angle of attack


text:
How much of an airplane's forward energy is lost to lift?
An airplane needs to move forward to generate lift, and because energy isn't created from nothing, all the kinetic energy of lift comes in the form of drag, where air (air resistance) turns forward motion into upward motion.
Some of the drag on an airplane is therefore "productive" in that it produces lift, and some is "unproductive" in that it takes forward motion from the plane without providing lift.
What percentage of an airplane's forward kinetic energy loss due to drag is productive as opposed to unproductive?

It appears you are asking what percent of drag is "induced drag", the word for drag due to the creation of lift.  Expect to see some graphs in the answers--  .  May not  be correct to equate drag with kinetic energy loss unless you want to specify a particular thought experiment like thrust is zero and pilot is manipulating controls to maintain altitude as airspeed bleeds off.  So you could improve question by deleting reference to KE if you can get to it before an answer is offered.

Is there any energy lost to lift?  It costs a boat nothing to maintain altitude.  Maybe once wings are going a certain speed, it costs them nothing to maintain altitude.  Would a different wing shape (with no lift) actually have less drag?

There is both induced drag, and parasitic drag.  See sections 1.2.5,  2.13.8 and 4.5 here in this great e-book "See How It Flies" https://www.av8n.com/how/

@Fattie Yes, wing with no lift costs less drag. If a wing can maintain lift without drag penalty, then it's a perpetual motion machine.

Can we please find a better "duplicate" question than that? This question is absolutely conceptual, while that question & its answer are entirely math & formulas. I'd almost vote to close as too broad, but a good "here's why it depends" answer could really work. Surely we have one of those somewhere? In a form accessible to someone who's asking this type of question?

@Jimmy , hmm, all bodies move indefinitely (inertia).  all bodies (say, the Earth, or my Beagle) have drag.  it's not immediately clear that the "lift related part" of a wing .. adds drag over the "ordinary everyday drag" of the wing.  it's a tricky issue.  an extremely light aircraft, what about a hot air balloon, will move around forever until the universe decays, it has "no lift drag" in the sense of the question, but it has "ordinary" drag (and, that's not related to perpetual motion).  It's tricky!

You almost never use kinetic energy (as best as I can understand “forward” in this context) to produce lift, because that would mean you are slowing down and you can only do that to a very limited point. Either you are using potential energy, if you are gliding, or you are using chemical energy (“dead dinosaurs”) from the fuel, using the engine to do the conversion.

@Fattie A hot air balloon naturally wants to go higher due to its bouyancy; bringing it down now requires extra energy. Plane is heavier than air so you're comparing apples to oranges. Nothing moves indefinitely unless it's in true vacuum (which does not exist).

I don't think we can really say that "all the kinetic energy of lift comes from drag".  Seems like an apples - to - oranges comparison.  Units of energy are different from units of drag.

In the most simple model for subsonic aerodynamics, drag is split into two components:

Zero-lift drag, that is all the drag created when the airplane produces no net lift. This kind of drag has again two components: Friction and pressure drag, that is the aerodynamic drag parallel and perpendicular to the local surface. This drag would dominate in a vertical dive or a zero-g parabola.
Drag created due to lift. Since that was explained mathematically first by using the Biot-Savart law for electromagnetic induction, this is called induced drag. The simplest explanation is: Lift is created by bending the oncoming air slightly downwards, and the reaction force is perpendicular to the mean angle of that airstream. Induced drag is the force component parallel to the initial direction of motion of the air relative to the airplane, and lift is the perpendicular component of that force. Thus, induced drag is lift times half the tangent of the bending angle.

While zero-lift drag increases with dynamic pressure, i.e. with the square of airspeed times density, induced drag decreases with dynamic pressure. Like this:

Drag components over speed for a typical glider (own work). The nonlinearity at the lowest speed is due to flow separation when the lift-creating limits of the aircraft are approached. The physics for large aircraft are the same, only the numbers will be larger.
Due to the dependency on the square of airspeed, the sum of both components has a minimum when they are of equal magnitude. However, given enough thrust, a motorized aircraft can sustain level flight at the far right end of that diagram when lift-dependent drag almost vanishes.

Interesting to note: proportionally lift induced drag almost vanishes (compared to form drag), but in terms of force units, it may actually increase if wing (and fuselage) AoA is less than optimal L/D AoA at higher airspeeds.

@RobertDiGiovanni That is right, but do not forget that the lift coefficient goes down linearly with dynamic pressure but the induced drag coefficient goes down with the square of the lift coefficient. So absolute induced drag is proportional to the lift coefficient.

When flying at the airspeed that yields the maximum L/D ratio, which is also the airspeed that yields the lowest total drag force, 50% of the total drag is "induced drag", i.e. drag due to the creation of lift.  At higher airspeeds, a lower % of the total drag is "induced drag".  At lower airspeeds, a higher % of the total drag is "induced drag".

Ps as an aside, the airspeed for minimum drag referenced in my answer, is NOT the airspeed for minimum drag coefficient.

By definition, energy is transferred to an aircraft via the application of thrust along a displacement. For the sake of simplicity we consider a displacement unitary and energy is therefore just proportional to thrust.
In level flight, thrust simply equals drag and therefore the question can be rephrased as: how much changes drag when lift goes to zero, i.e. how much changes $C_D$ @ $C_{L=0}$.
Those values are plotted on the polar of the aircraft, like this one for a B747-100 taken from ¹:

So, from this polar it's easy to see that $C_D$ @ $C_{L=0}$ is some 0.019 at Mach 0.86.
This value has to be compared with the drag when the lift is actually being produced. Let's consider then a standard cruise condition at height of 10km, Mach as before and mass of 250,000kg (halfway between MTOW and OEW for a B747-100). This cruise condition gives a $C_L=0.38$ and from the polar $C_D=0.023$.
The difference with the previous value is $0.023-0.019=0.004$ which corresponds (at the cruise condition just given) to a drag due to lift of $28kN$. This value multiplied by the displacement (in $m$) is the energy lost to lift we were looking for. For a range of say $8'500km$ that gives an energy lost to lift of some $240GJ$.
Two side notes: obviously during cruise height, speed, Mach and weight change and also the 0.004 that we have calculated changes as well; anyway the order of magnitude for the energy is correct. Even if the energy lost to lift look big, it is actually only 0.004 of the original 0.023 i.e. it corresponds only to the 17% of the whole energy used to keep the jumbo jet in flight at 10km and Mach 0.87.

¹ Dr. Jan Roskam, Airplane Design Part VI, DARcorporation

No energy is 'lost' to lift. All the energy delivered by the engine is spent, directly or indirectly, in accelerating air downwards in order to produce lift...

no, only half the energy. The rest is used to heat the boundary layer.

Simply BY DEFINITION, no energy can be lost to "Lift".  Lift is the portion of all aerodynamic force on the aircraft that lies normal (perpendicular) to the flight path motion of the aircraft. So being perpendicular to the flight path, it can neither accelerate nor decelerate the aircraft.
Drag is the portion of the total aerodynamic force that lies parallel to the direction opf motion. So, simply by definition, it is the drag that slows the aircraft.
But your question (restated as "How much of an airplane's forward energy is lost to Aerodynamic force?" is a good one, and it is dependent on the Angle of Attack, (AOA). The higher the AOA, the more the total Aerodynamic force vector is tilled aft (backwards), from the aircraft direction of motion. This means that as AOA increases, the Drag component of that force (as a percentage of the total) will increase, and the Lift component percentage will decrease.

The surprise of some engineers about comparing kinetic energy to lift?!?.
Yes, let's look at the formulas, and how we apply them.  In steady state flight, we talk about the 4 FORCES of flight. F = kg m/s$^2$ = ma:  Lift, Gravity, Thrust, Drag
So why not Kinetic Energy  KE = kg m$^2/s^2$ = F × d, or even Power P = kg m$^2/s^3$ = F × d/t?
Because, in describing forces on an aircraft in steady state flight, everything happens in the same distance at the same time, so d/t cancels out, leaving a balance of forces.
However, we can easily determine total drag from the glide ratio.
The energy state of an aircraft is its potential energy mgh + its kinetic energy 1/2mv$^2$.
Assuming it maintains constant speed in a glide and flys in a straight line at constant AOA to its landing point, energy consumed is all mgh: kg m$^2$/s$^2$.
There for total drag force × distance = mgh
So if the glide ratio is 10 to 1, total kinetic energy converted from potential energy is around kg m/s$^2$ × distance = kg × gravity (m/s$^2$) × height.  Canceling units we have,
for a 1000 lb airplane (thrust = drag): 100 lbs of thrust force is required for level flight.
To determine lift to drag ratios, we move to the wind tunnel, and develop data points for
the formula Ctotal drag = Cdrag form + Cdrag induced.  As seen in Peter Kampf's graph, this will vary depending on airspeed and angle of attack.
A better wing will have a higher lift to total drag ratio, and considering that these are aircraft, a better question may be how much "forward energy" is lost NOT creating lift.
Drag is a product of motion, and apportioned to lifting AND moving forward, nothing is wasted or lost.  "Forward energy" is thrust × distance.  What one can do next is to begin to work on efficiency, so one may need less to go as high and as far.

-1 because I can barely read the math :( Please use Mathjax to make it more understandable.

Revived with Mathjax corrections.

About 6%
You're looking for the "Lift to Drag" ratio - see here for some examples of different things: https://en.wikipedia.org/wiki/Lift-to-drag_ratio
A paraglider is a 10:1 glide ratio.
A 747 at cruise is 17:1  - thus - 6% of the power keeps it up, and the rest, 94%, offsets the drag of flying at mach 0.85.

No way.  When a high-performance sailplane is cruising at the max L/D ratio, 50% of the drag is induced drag.  PS this could be the basis of an answer.

I don't think this is what was asked. When you have L/D = 17/1, this means you just need 6% of thrust with respect to lift (and weight). The total 'energy supplied' is commensurate with this amount. Now the question is, how much of these 6% is 'useful' (needed for lift) and how much just a 'waste'.

That's not what lift-to-drag ratio mean. Lift-to-drag ratio 17 means that it only needs thrust equal 1/17 (~6%) of lift (which is equal to weight) for stable horizontal flight.


text:
Please explain real-world procedures behind this "ATC joke"
One of the well-known kind of the "ATC jokes" (funny excerpts from real or invented communication between pilots and air traffic controllers) has recently caught my eye.
I can't seem to understand what actual procedures or rules does it refer (or allude) to:

Pilot: "Approach, Federated 303's with at 8000' for vectors ILS, full stop.
Approach: "Unable Federated 303. The ILS is out of service."
Pilot: "We'll take the VOR then."
Approach: "Sir, the VOR's in alarm right now. Standby."
Pilot: "OK, guess it'll have to be the ADF then."
Approach: "303, unable the ADF right now for traffic saturation."
Pilot: "OK, approach. State my intentions."

(src)
Could you please enlighten me on what does the last (highlighted) line is intended to mean and/or convey? I understand it's a word play on "state your intentions", but what exactly is it supposed to mean?

The lack of request for GPSS approach marks out just how old this joke is.

@Jamiec: Not old enough for them to be requesting a PAR approach, though.

As you commented, it's a play on "state [say] your intentions". However, the pilot has said his intentions several times and ATC rejected them each time, so he's turning it around by asking ATC to tell him what intentions won't be rejected.
The "intentions" in this case are the type of Instrument Approach Procedure (IAP) he wants to use. Typically there are several for each runway, for use by aircraft with different levels of equipment; ILS, VOR and ADF (actually NDB) are common options in order of decreasing desirability.
Depending on the weather, there may be two additional IAPs available at any airport which aren't usually charted: visual and contact. There's a possible second-level joke here in that ATC isn't allowed to offer a contact approach, but the pilot is allowed to request it--so if the pilot is asking ATC to tell him his intentions, then in the logic of the joke, ATC could tell him he's requesting that. OTOH, if the weather were such that a visual approach were available, ATC probably would have assigned it to him already without any requests.

"VOR's in alarm" is that a permanent problem.

@user2617804 I believe it means the VOR system is broken, and some monitoring system has detected that it is broken (raised an alarm saying it's broken). (Not a pilot or ATC)

It's an ironic word play on the common ATC request "state your intentions" when a follow-on action is at the pilot's discretion and ATC needs to know what the plan is.
Most large airport runways have several types of instrument approaches, the most precise being ILS, and down from there. It means they've gone down a list of approach options from most precise to least precise (ILS, VOR, ADF) and there aren't any more, so the pilot is saying what amounts to "I'm out of options, so tell me what to do next".
Although the specific phrases being used there are kind of non-sensical, as if written by a non-pilot, it conveys the gist of it for the purposes of the gag.

I understand it's a word play on "state your intentions", but what exactly is it supposed to mean? That the pilot asks/expects the ATC to tell them what to do because they've run out of options to consider?


text:
Is extreme wind shear an official classification? Is it rare to land in 65kts headwind, gusting 95kts?
I was a passenger in an A340 during a famous storm 65kts gusting 95kts. Resulted in a go around, but the plane sunk quite fast several times. Is this rare? Are there limits to headwind component? The captain said we experienced extreme wind shear, is that an official grade? Checked the metar for the kts.

Hi Tapani, welcome to aviation.SE. Please be aware that we prefer to have only one question per post, and you currently have 4-5 (depending on how you count). Can you please focus your post on one single question? Otherwise it risks being closed as too broad. thank you.

Some specifics would be nice: where and when this happened?

Wind shear is also gusts, but the difference is that wind shear involves changes in speed and direction, where gusts are generally in the same direction. 65kt gusting 95kt is a very rare situation that no airline captain would want to tackle unless the situation demands it (low fuel, some other emergency) and the winds are straight as an arrow down the runway. You could land the airplane that way, but a 95 kt gust can certainly lift a wing or cause weathervaning when the aircraft is taxiing off the runway.
Now, if the report is for wind shear gusting that high, it would be crazy to try to land an A340 (or any other aircraft for that matter) unless the pilot had zero other choices (or just didn't know that there was low-altitude wind shear). The max crosswind component for landing on an A340 is around 37 kts.
From what you are describing, it seems they experienced wind shear, and not just gusts.

Isn't wind shear defined as an unusually rapid change in wind speed / direction as a function of altitude?

@pericynthion Wind shear can be measured horizontally or vertically but that is not a requirement to call it such.


text:
What's the purpose of the holes on the Cabri G2 fenestron blades?
What is the purpose of the holes on the blades of the Cabri G2 fenestron? In my opinion drilling the blades like that weakens their cross section under the action of the centrifugal force. Is it something related with the manufacturing process or with the blades aerodynamics and boundary layer?

I think that the holes you are mentioning are placed on the hub and they are bigger because the straps are located bellow the blades roots. The holes through each of the blade are very small and I'm not sure what you can inspect from them.

It's the aviation equivalent of whistle tips Woo Wooo!

@MichaelHall The text in the article is misleading. As I suspected the holes for straps inspection are on the hub - check link and link . The reason for the small holes drilled through every blade still stays unknown.

I stand corrected, thanks.

Are the holes only present for fenestron blades?

If changing the angle of attack moves plastic skin relatively to steel core and if a gap is created, holes could be for air pressure equalization.

The Cabri G2 has "fail-safe injected plastic blades" (pdf) on the tail rotor. Those holes might simply be the injection points.

@sophit They do not look like gating points, but I think mins has a credible idea: If those blades are hollow and the hub is solid, those holes are for pressure equalisation. Two for redundancy, and the size need not be larger than they are.

@PeterKämpf: I can count at least seven of them on each blade

@sophit OK, did not check that picture. But that makes a gating point even less likely, as only one is needed (and best for avoiding junctions). But then, more than two equalisation holes would also not be needed.

These holes come from the manufacturing process. They inject plastic into the blade mould through these holes. The holes don't serve any aerodynamic purpose. That's the explanation I got at a recent visit to the plant in Aix-en-Provence.

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